Problem: On the way to school, Karen became fixated on the puzzle game on her phone!

<image>

The game is played as follows. In each level, you have a grid with n rows and m columns. Each cell originally contains the number 0.

One move consists of choosing one row or column, and adding 1 to all of the cells in that row or column.

To win the level, after all the moves, the number in the cell at the i-th row and j-th column should be equal to gi, j.

Karen is stuck on one level, and wants to know a way to beat this level using the minimum number of moves. Please, help her with this task!

Input

The first line of input contains two integers, n and m (1 ≤ n, m ≤ 100), the number of rows and the number of columns in the grid, respectively.

The next n lines each contain m integers. In particular, the j-th integer in the i-th of these rows contains gi, j (0 ≤ gi, j ≤ 500).

Output

If there is an error and it is actually not possible to beat the level, output a single integer -1.

Otherwise, on the first line, output a single integer k, the minimum number of moves necessary to beat the level.

The next k lines should each contain one of the following, describing the moves in the order they must be done:

  * row x, (1 ≤ x ≤ n) describing a move of the form "choose the x-th row". 
  * col x, (1 ≤ x ≤ m) describing a move of the form "choose the x-th column". 



If there are multiple optimal solutions, output any one of them.

Examples

Input

3 5
2 2 2 3 2
0 0 0 1 0
1 1 1 2 1


Output

4
row 1
row 1
col 4
row 3


Input

3 3
0 0 0
0 1 0
0 0 0


Output

-1


Input

3 3
1 1 1
1 1 1
1 1 1


Output

3
row 1
row 2
row 3

Note

In the first test case, Karen has a grid with 3 rows and 5 columns. She can perform the following 4 moves to beat the level:

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In the second test case, Karen has a grid with 3 rows and 3 columns. It is clear that it is impossible to beat the level; performing any move will create three 1s on the grid, but it is required to only have one 1 in the center.

In the third test case, Karen has a grid with 3 rows and 3 columns. She can perform the following 3 moves to beat the level:

<image>

Note that this is not the only solution; another solution, among others, is col 1, col 2, col 3.

Solution (in JAVA):

Solution:
import java.lang.reflect.Array;
import java.math.BigInteger;
import java.util.*;


public class Main {

    public static void main(String[] args) {
        Scanner scan = new Scanner(System.in);
        int n = scan.nextInt();
        int m = scan.nextInt();

        int[][] g = new int[n][m];

        for (int i = 0; i < n; i++)
        {
            for (int j = 0; j < m; j++) {
                g[i][j] = scan.nextInt();
            }
        }

        int[] c = new int[m];
        int[] r = new int[n];

        int minFound = Integer.MAX_VALUE;
        int[] c_ans  = new int[0];
        int[] r_ans = new int[0];

        for (int f = 0; f <= g[0][0]; f++) {
            r[0] = f;
            c[0] = g[0][0] - f;

            boolean isCorrect = true;
            for (int i = 1; i < n; i++) {
                r[i] = g[i][0] - c[0];
                if (r[i] < 0)
                    isCorrect = false;
            }

            for (int i =1; i < m; i++) {
                c[i] = g[0][i] - r[0];
                if (c[i] < 0)
                    isCorrect = false;
            }
            if (!isCorrect)
                continue;

            for (int i = 0; i < n; i++) {
                for (int j = 0; j <m; j++) {
                    if (r[i] + c[j] != g[i][j]) {
                        isCorrect = false;
                        break;
                    }
                }
                if (!isCorrect){
                    break;
                }
            }

            if (isCorrect) {
                int sum = 0;
                for (int i = 0; i < n; i++)
                    sum += r[i];

                for (int j = 0; j < m; j++)
                    sum += c[j];

                if (minFound > sum) {
                    minFound = sum;
                    c_ans = c.clone();
                    r_ans = r.clone();
                }


            }
        }

        if (minFound < Integer.MAX_VALUE) {
            System.out.println(minFound);

        for (int i = 0; i < n; i++) {
            for (int e = 0; e < r_ans[i]; e++) {
                System.out.println("row " + (i + 1));
            }
        }

        for (int j = 0; j < m; j++) {
            for (int e = 0; e < c_ans[j]; e++) {
                System.out.println("col " + (j + 1));
            }
        }
        }

        else
            System.out.println("-1");


    }
}